Forum:Iteror.org?
Has anyone visited the site iteror.org? For example, at http://iteror.org/big/Work/royal/planB.html they claim to devise a notation to beat Bird's H(n). (not sure if they are talking abou his old version or his new version) But I can't make heads or tails of their notation. What do you guys think? Deedlit11 (talk) 22:03, October 30, 2013 (UTC) :The use of generalized regular expressions seems interesting. I don't quite understand everything yet, but I'll be looking over it for a while. FB100Z • talk • 22:35, October 30, 2013 (UTC) :Okay, so skimming through the article gives me the impression that Lelieveld has built a system about as powerful as BEAF up to legion arrays. Instead of representing things as arrays, all the manipulations occur syntactically, so you'd write 111,,11,1111 instead of {3,0,2,4}. That explains why the notation is so alien. FB100Z • talk • 23:07, October 30, 2013 (UTC) :i am Necroing this thread but i want to know exactly how powerful beatrix is 22:30, March 28, 2014 (UTC) ::I've started a section below so we can collaboratively decode the page. FB100Z • talk • 20:49, March 29, 2014 (UTC) :::Thanks for this. Unfortunately, I'm not sure how much I can help since the page still confuses the heck out of me. How far do you think the notation goes again? Although I can't grok the notation, it looks like the standard procedure up to \(\varepsilon_0\); however, there is the part about "dual deeps" and "serial deeps". I'm not sure how much that extends the notation, if at all. :::Also, have you taken a look at the rest of the site? Deedlit11 (talk) 11:40, March 30, 2014 (UTC) :::::The rest of the site looks also interesting, they say they have early evidence for tetration. Also, I don't think BTRIX is really what we're expecting. It seems it reaches something around LVO and it then implents Bird's Array in a different system. The page also says that they can always beat other notations, with ther RepExp notation and such things. Wythagoras (talk) 18:53, April 11, 2014 (UTC) ::::RepExp is, of course, only a small fraction of the page. BTRIX analysis is coming soon. As for the strength of BTRIX, it really just looks like a slightly modified clone of BEAF; Lelieveld takes a lot of time to boast about how his* notation outruns the latter, but I'm skeptical about how substantial his advantage is. ::::Dunno about the rest of the site, but if it's all as cryptic as this page, I'm not sure if I want to work through it! ::::Gerard is a male name, right? FB100Z • talk • 19:13, March 30, 2014 (UTC) ::::Yes, it is. Wythagoras (talk) 19:18, March 30, 2014 (UTC) Huh, why do the subscripts keep automatically breaking with every edit? FB100Z • talk • 19:22, March 30, 2014 (UTC) Aw, what happened to this? King2218 (talk) 15:05, September 21, 2014 (UTC) Stuff we know so far RepExp Lelieveld has created a system called "RepExp" (short for "repeated expressions"), which he calls a meta-notation for describing recursive notations. RepExp differs from existing notations in that it operates on syntactic expressions rather than argument spaces. No formal definition is given for RepExp, but the following examples are offered: x.. :3 = x.. x:3 = x{3} = xxx F(..m..) :3: = F(F(F(m))) A.n1,..Z :2 = A(n1,){2}Z = An1,n1,Z aaaa..b..b.ci.. aa:0 2: :4 = aabbbc0c1c2c3 A.Li..,{5}..Rj.Z :3: = AL0L1L2,,,,,R2R1R0Z From the first and last examples, we see that a{n} represents repetition of a'' ''n times: x{5} = xxxxx. The third example hints that this curly-brace notation by default only repeats one character, and to repeat multiple characters we use parentheses (ab){n}: (abc){3} = abcabcabc. This is no surprise to people familiar with regular expressions, which Lelieveld cites as inspiration. But this notation is not as important as the dot notation. Dot notation strings appear to have two parts: a main body and a sort of appendix that indicates the number of repetitions. The main body contains single dots and double dots that, in some cryptic way, delimit the expression or expressions to be repeated. Here is a pseudocode formalization of what I've figured out: *The main body consists of string with single and double dots interspersed. *The appendix consists of one or more words of the form ":n", "n:", "A:n", "n:A", or ":n:", where A is a non-empty string and n is a nonnegative integer. *A word of the form ":n:" (tandem repetition) is shorthand for ":n n:". *After having expanded all the tandem repetition words, we match up each appendix word to a successive double-dot. For a valid expression, the number of words and dot-pairs must match (again, with each tandem repetition counting as two words). *For each appendix word: **If it is ":n", start at the corresponding double-dot and walk backwards in the string until you reach a single dot or the beginning of the string. Take the string that you've walked over, trim off the delimiting double-dot, and repeat that string n times. **If it is "n:", start at the corresponding double-dot and walk forwards in the string until you reach a single dot or the end of the string. Take the string that you've walked over, trim off the delimiting double-dot, and repeat that string n times. **If it is "A:n", there must be an occurrence of string A immediately before the corresponding double-dot. (If not, the expression is invalid.) After removing the double-dots, repeat that occurrence of A n times. **If it is "n:A", there must be an occurrence of string A immediately after the corresponding double-dot. (If not, the expression is invalid.) After removing the double-dots, repeat that occurrence of A n times. *Remove any excess single dots. For example, let's try aaaa..b..b.c.. aa:0 2: :4, which is almost the same as the fourth example. First we pair each appendix word with each pair of double-dots: aaaa..b..b.c.. ^^ ^^ ^^ aa:0 2: :4 First we look at aa:0. The string "aa" does appear right before the double-dot, so we take that and repeat it zero times. In other words, we delete it: aab..b.c.. ^^ ^^ 2: :4 Note that we deleted only the second "aa", and the first remains untouched. Next is 2:, which selects up to the next single-dot, selecting the second "b". Repeating that twice: aabbb.c.. ^^ :4 :4 walks backwards to the preceding single-dot, which selects the c and repeats it four times: aabbb.cccc Finally, we remove the extraneous single-dot to get: aabbbcccc RepExp + subscripts The last two examples contain a subscript notation not covered by the above definition. So let's fix that: *The main body consists of a string with single and double dots, and zero or more subscripts which may be letters or numbers. *The appendix consists of one or more words of the form ":n", "n:", "A:n", "n:A", or ":n:", where A is a non-empty string and n is a nonnegative integer. *A word of the form ":n:" (tandem repetition) is shorthand for ":n n:". *After having expanded all the tandem repetition words, we match up each appendix word to a successive double-dot. For a valid expression, the number of words and dot-pairs must match (again, with each tandem repetition counting as two words). *For a string S'', a nonnegative integer ''n, and a direction d'', define repeat(''S, n'', ''d) as follows: **If n'' = 0, return an empty string. **In ''S, replace any letter subscript with a subscript n'' - 1 to form a string ''T. **If d'' = left, return repeat(''S, n'' - 1, ''d) + T''. **If ''d = right, return T'' + repeat(''S, n'' - 1, ''d). *For each appendix word: **If it is ":n", start at the corresponding double-dot and walk backwards in the string until you reach a single dot or the beginning of the string. Take the string that you've walked over and call it S''. Trim off the delimiting double-dot, and replace ''S with repeat(S'', ''n, left). **If it is "n:", start at the corresponding double-dot and walk forwards in the string until you reach a single dot or the end of the string. Take the string that you've walked over and call it S''. Trim off the delimiting double-dot, and replace ''S with repeat(S'', ''n, right). **If it is "A:n", there must be an occurrence of string A immediately before the corresponding double-dot. (If not, the expression is invalid.) Call that occurrence of A S''. Trim off the delimiting double-dot, and replace ''S with repeat(S'', ''n, left). **If it is "n:A", there must be an occurrence of string A immediately after the corresponding double-dot. (If not, the expression is invalid.) Call that occurrence of A S''. Trim off the delimiting double-dot, and replace ''S with repeat(S'', ''n, right). *Remove any excess single dots. This notation is quite complicated for something explained in two short paragraphs and five examples. However, Lelieveld's examples are comprehensive enough to understand almost all of his uses of the notation throughout the rest of the page. You actually don't need to work through this definition to get it, but I've written it up because formalize formalize formalize. Although I was skeptical at first, I find RepExp actually quite a nice abbreviation after having worked it out. f(2, f(2, f(2, f(2, 3)))) can be nicely put as f(2,..3..) :4:, for example, and I can't actually think of a more succinct way of writing it! BTRIX: overview It is important to note that BTRIX is a syntactically based system. The BTRIX incarnation of a {3, 4, 5, 6} linear array is :111,1111,11111,111111 I don't know how much of an advantage this gives, although I can see how it can create very clear rules when we get to complex nested dimensional things. Lelieveld uses A = B to mean that string A reduces to string B in the BTRIX system, and A ≡ B to mean that string A and string B are literally the same. I have mimicked this fairly sensible notation here as well. BTRIX: Linear arrays In BTRIX, we have actual rules to deal with, thankfully. Here's linear array notation: :0.0: BTRIX(X) ≡ X :0.1: BTRIX(a) ≡ a = a :1.0: a, = string :1.1: a,b = b :2.1: a,Y, = a,Y :3.0: a,b,1c = a,ab,c :3.1: a,b,1c,d = a,ab,c,d :4.0: a,,,1d = a,a,,1d :5.0: a,b1,,1d = a,,b1,1d :3.2: a,b,1Z = a,ab,Z :4.1: a,,{k1}1Z = a,a,{k1}Z :5.1: a,b1,{k1},1Z = a,,{k1}1b,Z where lowercase letters represent a string of 1's, and uppercase letters represent strings of 1s and commas. It appears that the rules are structured so that, say, 3.2 is a generalization of 3.1, which is a generalization of 3.0. Thus we only need the following: :0.1: BTRIX(a) ≡ a = a :1.1: a,b = b :2.1: a,Y, = a,Y :3.2: a,b,1Z = a,ab,Z ("motor") :4.1: a,,{k1}1Z = a,a,{k1}Z ("put") :5.1: a,b1,{k1},1Z = a,,{k1}1b,Z ("upload") From this, we can gather that BTRIX (so far) is a linear array notation with the following rules, expressed in Bowerslike form: * Zeros are default. * Let b'' (base) be theo value of first entry, and ''p (prime) be the value of the second entry. Define the pilot as the first nonzero entry after the prime, and the copilot as the entry before that. * If there is no pilot, set the base to p'' and the pilot to 0. Return. * If the pilot is the third entry, decrement it and add ''b to the prime. Return. * If the prime is 0, decrement the pilot and add b'' to the prime. Return. * Otherwise, decrement the pilot, set the copilot to ''p, and set the prime to 0. Return. How powerful is this? Linear BTRIX emulates the hyperoperators exactly, as noted by the author. Up to here the power is \(f_\omega\). BTRIX: Dimensional arrays The BTRIX 1-dimensional separator is denoted as ,1, the 2-dimensional separator as ,11, etc. It is understood that , is a shorthand for ,[]. :6.0: a,b,11Z = a,a,{b}1,1Z :6.1: a,b1.,1..1Z :k1 = a,a.,1..,{b1}1,1Z :k :2.2: a,Y,s = a,Y :3.2: a,b,1Z = a,ab,Z :4.2: a,.,si..1Z :k1 = a,a.,si..Z :k1 :5.2: a,b1.,si..,1Z :k1 = a,.,si..b1,Z :k1 :6.2: a,b1$,t11Z = a,a$.,t..1,t1Z :b1 where $ ≡ ,si.. :k by 8. {sk≥t1} :7.1: ,[] ≡ , :8.0: ,So the current ruleset is: :0.1: BTRIX(a) ≡ a = a :1.1: a,b = b :2.2: a,Y,[s = a,Y :3.2: a,b,1Z = a,ab,Z :4.2: a,.,si..1Z :k1 = a,a.,si..Z :k1 :5.2: a,b1.,si..,1Z :k1 = a,.,si..b1,Z :k1 :6.2: a,b1$,t11Z = a,a$.,t..1,t1Z :b1 where $ ≡ ,si.. :k by 8. {sk≥t1} :7.1: ,[] ≡ , :8.0: ,:That's awful, is it reach the level of \psi(\psi_I(0)) ? [[User:Ikosarakt1|Ikosarakt1] (talk ^ ) 14:28, March 31, 2014 (UTC) ::It's not as bad as it looks. No idea about the strength of the entire function, and I've only covered dimensional arrays. you're.so. 06:22, April 11, 2014 (UTC) In plain English: * Zeros are default. * Let b'' (base) be theo value of first entry, and ''p (prime) be the value of the second entry. Define the pilot as the first nonzero entry after the prime, and the copilot as the entry before that. * If there is no pilot, set the base to p'' and the pilot to 0. Return. * If the pilot is the third entry, decrement it and add ''b to the prime. Return. * If the prime is 0, decrement the pilot and add b'' to the prime. Return. * If there is a copilot (that is, the pilot does no sit at the beginning of a row), decrement the pilot, set the copilot to ''p, and set the prime to 0. Return. * Otherwise... ** Decrement the pilot. ** Let d denote the dimension of the smallest previous structure to the pilot. (d = 1 if the pilot sits on a row other than the first row of a plane, d = 2 if the pilot sits on a plane other than the first plane in a realm, etc.) ** uh, I don't know how to do this.